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        riddles >> medium >> prove  sec40° + sec80° + sec160° =6
        
(Message started by: tony123 on Aug 4^th, 2007, 9:35am)

Title: prove sec40° + sec80° + sec160° =6
Post by tony123 on Aug 4^th, 2007, 9:35am

prove
sec40° + sec80° + sec160° =6

// title changed --towr
(I'm not good at coming up with titles either, but when the entire problem statement fits there, why not!)

Title: Re: prove
Post by K Sengupta on Aug 5^th, 2007, 2:01am

At the outset, I would like to state that I found the fundamental tenets inclusive of the problem problem very interesting and I am positing a proposed solution in my subsequent post.

However, the problem title appears to be a duplication of a title corresponding to a previous contribution by the same author . In my opinion this is at direct variance with the provisions governing the website.

Accordingly, I would request for moderator intervention to change the problem title.

Title: Re: prove
Post by K Sengupta on Aug 5^th, 2007, 2:03am

Method 1:

[hide]Substituting h = 40^o, 80^o, 160^o in turn, we observe that:
cos 3h =-1/2, for each of the foregoing values, and accordingly:
8 cos³h – 6 cos h + 1 =0

Therefore, cos 40, cos 80 and cos 160 are the roots of the cubic equation
8 y³ – 6 y + 1 =0; for y = cos h and so, their respective reciprocals sec 40, sec 80 and sec 160 correspond to the roots of the cubic equation :
y³ – 6y + 8 = 0, and therefore:
The sum of the roots of the cubic is 6.

Consequently:
[EDIT] sec 40^o + sec 80^o + sec 160^o = 6[/hide]

Title: Re: prove
Post by ThudanBlunder on Aug 5^th, 2007, 4:19am

on 08/05/07 at 02:01:22, K Sengupta wrote:

However, the problem title appears to be a duplication of a title corresponding to a previous contribution by the same author . In my opinion this is at direct variance with the provisions governing the website.
[/i]

Yes, it has happened more than once (http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi?board=riddles_medium;action=display;num=1184282061;start=2#2).

As for the problem, it would be more interesting and harder if he asked to solve secx + sec2x + sec4x = 6

The above property of http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif/9 is not on Mathworld (http://mathworld.wolfram.com/TrigonometryAnglesPi9.html) (although Morrie's Law (http://mathworld.wolfram.com/MorriesLaw.html) is).

Title: Re: prove
Post by JocK on Aug 5^th, 2007, 5:02am

Cute problem.

on 08/05/07 at 04:19:19, ThudanBlunder wrote:

As for the problem, it would be more interesting and harder if he asked to solve secx + sec2x + sec4x = 6

Thinking about it, I'd like the formulation:

Prove that the three numbers cos40°, cos80° and cos160° have arithmetic mean 0, harmonic mean 1/2 and geometric mean -1/2.

Title: Re: prove sec40° + sec80° + sec160° =6
Post by tony123 on Aug 5^th, 2007, 2:28pm

K Sengupta

Method 1:

i think its rong
and
cos 40 + cos 80 + cos 160 not= 6

the problem is
sec40° + sec80° + sec160° =6

Title: Re: prove sec40° + sec80° + sec160° =6
Post by JocK on Aug 5^th, 2007, 2:44pm

on 08/05/07 at 14:28:41, tony123 wrote:

K Sengupta

Method 1:

i think its rong

Actualy, Sengupta is right: the stated follows from applying Viete's formulas to the cubic equation in cos(x) resulting from expressing 2cos(3x)+1=0 in terms of cos(x). It's just that in the very last line of his post a triple typo ('cos' instead of 'sec') pops up.

Title: Re: prove sec40° + sec80° + sec160° =6
Post by K Sengupta on Aug 6^th, 2007, 8:32am

on 08/05/07 at 14:44:06, JocK wrote:

It's just that in the very last line of his post a triple typo ('cos' instead of 'sec') pops up.

True.

I was careless and made the typo of triple cos instead of sec.

I confirm having corrected the foregoing anomaly.

Title: Re: prove sec40° + sec80° + sec160° =6
Post by K Sengupta on Aug 6^th, 2007, 8:56am

I furnish hereunder, a second proposed method giving the solution to the problem.

I am not certain if this corresponds to the author's official solution.

Method 2
[hide]sec 40 + sec 80 + sec 160

= (cos 40 + cos 80)/(cos 40*cos80) + 1/(cos 160)

= (2*cos 60*cos 20)/(cos 40*cos 80) + 1/ (cos 160)

= (cos 20*cos 160 + cos 40*cos 80)/(cos 40*cos 80*cos 160)

= (cos 180 + cos 140 + cos 120 + cos 40)/ {cos 40*(cos 240 + cos 80)}

= (-1 - 1/2)/{(1/2)(-cos 40 + cos 120 + cos 40)}

= (-3/2)/(-1/4)

= 6[/hide]